Why Interior Candidates Matter
To find the absolute maximum or minimum of a function on a region, it is not enough to solve equations without regard to where their solutions lie. The region determines which points are possible. A point where both first partial derivatives vanish is relevant as an interior candidate only if it actually lies inside the region.
The basic procedure is simple to state: solve \(f_x(x,y)=0\) and \(f_y(x,y)=0\) simultaneously, then filter the solutions using the region. The simultaneous solution matters because a function of two variables can change as either coordinate changes.
A stationary point is a candidate, not a guaranteed maximum or minimum. It could be a local extremum, a saddle point, or a point where further analysis is needed. Also, a stationary point outside \(D\) cannot be an interior candidate for optimization on \(D\), even if it solves both equations.
Why Both Partial Derivatives Must Vanish
At an interior local maximum or minimum, small changes in either coordinate are allowed. Holding \(y\) fixed and changing \(x\) gives a one-variable function; holding \(x\) fixed and changing \(y\) gives another. At a local extremum, each of these one-variable functions has a local extremum, so its derivative must be zero.
Here is the argument in detail. Because \((a,b)\) is an interior point of the domain, there is a small open disk around it contained in the domain. In particular, for sufficiently small \(h\), both \((a+h,b)\) and \((a,b+h)\) lie in that disk. Define the coordinate slices \(g(h)=f(a+h,b)\) and \(q(h)=f(a,b+h)\). If \(f\) has a local maximum or minimum at \((a,b)\), then \(g\) has a local maximum or minimum at \(h=0\), and so does \(q\). By the one-variable necessary condition for a differentiable local extremum, \(g'(0)=0\) and \(q'(0)=0\). The chain rule gives \(g'(0)=f_x(a,b)\) and \(q'(0)=f_y(a,b)\). Therefore both partial derivatives vanish.
This is a necessary condition, not a sufficient one: the equations identify points that must be checked, but do not establish that an extremum occurs there. For absolute extrema on a compact region, the candidate principle also requires considering the boundary. The present procedure finds the interior candidates; it does not replace the boundary analysis.
A Reliable Procedure
Differentiate with respect to \(x\) while treating \(y\) as constant, then with respect to \(y\) while treating \(x\) as constant.
Solve the simultaneous system \(f_x(x,y)=0\), \(f_y(x,y)=0\). Do not solve just one equation.
Substitute each solution into both equations. This catches algebraic errors and confirms that it is stationary.
Retain only solutions that are strictly inside the region. A solution outside the region is not an interior candidate for this problem.
Worked Examples
Worked Example: A Single Interior Candidate
Find the interior stationary candidates of \(f(x,y)=x^2+y^2-4x+2y\) on the rectangle \(D=\{(x,y):0\leq x\leq 3,\ -2\leq y\leq 1\}\).
First compute the partial derivatives:
Set both equal to zero. The first equation gives \(2x-4=0\), so \(2x=4\) and \(x=2\). The second gives \(2y+2=0\), so \(2y=-2\) and \(y=-1\). Thus the only solution of the system is \((2,-1)\).
Check it in both equations: \(f_x(2,-1)=2(2)-4=4-4=0\), and \(f_y(2,-1)=2(-1)+2=-2+2=0\). The rectangle conditions are \(0\leq x\leq3\) and \(-2\leq y\leq1\). At this point, \(0\leq2\leq3\) and \(-2\leq-1\leq1\); both coordinates are strictly between their endpoints. Therefore \((2,-1)\) is an interior stationary candidate.
Its function value is \(f(2,-1)=2^2+(-1)^2-4(2)+2(-1)=4+1-8-2=-5\). This value is recorded for comparison with other candidates, but by itself it does not show that the point is an absolute extremum on \(D\).
Worked Example: Several Stationary Points, One Outside
Find the interior stationary candidates of \(f(x,y)=(x^2-1)^2+(y-1)^2\) on \(D=\{(x,y):-\frac12\leq x\leq\frac32,\ 0\leq y\leq2\}\).
Differentiate each term:
The first stationary equation factors as \(4x(x^2-1)=4x(x-1)(x+1)=0\), so \(x=-1\), \(x=0\), or \(x=1\). The second equation gives \(2(y-1)=0\), so \(y=1\). The full system therefore has exactly three solutions: \((-1,1)\), \((0,1)\), and \((1,1)\).
Verify the solutions: for each listed \(x\), one factor of \(4x(x-1)(x+1)\) is zero, and at \(y=1\), \(2(y-1)=0\). Now check the region. The point \((-1,1)\) is outside because \(-1<-\frac12\). The points \((0,1)\) and \((1,1)\) satisfy \(-\frac12<x<\frac32\) and \(0<y<2\), so both are interior stationary candidates.
Their values are \(f(0,1)=(0^2-1)^2+(1-1)^2=1+0=1\), and \(f(1,1)=(1^2-1)^2+(1-1)^2=0+0=0\). The excluded solution has value \(f(-1,1)=((-1)^2-1)^2+0=0\), but its value is irrelevant to the optimization on \(D\): the point is not in the region.
Worked Example: A Stationary Solution Outside the Region
Determine whether \(f(x,y)=xy\) has any interior stationary candidates on \(D=\{(x,y):1\leq x\leq2,\ -2\leq y\leq-1\}\).
The partial derivatives are \(f_x(x,y)=y\) and \(f_y(x,y)=x\). The simultaneous equations are \(y=0\) and \(x=0\), with unique solution \((0,0)\).
Substitution verifies the solution: \(f_x(0,0)=0\) and \(f_y(0,0)=0\). But every point of \(D\) has \(1\leq x\leq2\) and \(-2\leq y\leq-1\). The point \((0,0)\) satisfies neither coordinate condition, so it is outside \(D\). Therefore there are no interior stationary candidates in this region.
This conclusion does not mean the function has no absolute extrema on \(D\). It means only that no interior point of \(D\) solves the stationary equations. Any absolute extrema must be found among the other candidates required by the region, including its boundary.
Common Pitfalls
Setting only one partial derivative equal to zero. For example, if \(f(x,y)=x^2+y^2\), then \(f_x=2x\). Solving \(f_x=0\) gives every point with \(x=0\), including \((0,1)\). But \(f_y(0,1)=2\), not zero. The correct system is \(2x=0\) and \(2y=0\), whose only solution is \((0,0)\). A point must satisfy both equations.
Keeping a solution just because it solves the equations. In the second worked example, \((-1,1)\) solves both equations, but it lies outside the specified rectangle. It cannot be an interior candidate for that region. Always check the region after solving the system.
Calling every stationary point an extremum. The equations locate stationary points; they do not classify them. Even an interior stationary point may be a saddle point. Further analysis and comparison with the region’s other candidates are needed to determine absolute extrema.
Why the Region Check Is Essential
The stationary equations depend on the function, while the candidate list for an optimization problem depends on both the function and the region. The same function may have a stationary point in one region and no stationary points in another. That is why the process has two distinct stages: first solve the equations, then test membership in the region’s interior.
For a region described by inequalities, translate those inequalities into direct checks on each solution. To be an interior point, a solution must satisfy the region’s conditions with room to move in every direction locally; merely lying on an edge is not enough. Boundary points are handled separately when searching for absolute extrema.
Check Your Understanding
For each question, distinguish solving the stationary equations from deciding which solutions are relevant to the region.
- What two equations define a stationary point of a differentiable function \(f(x,y)\)?
- Why must an interior local maximum or minimum have both first partial derivatives equal to zero?
- If a solution of the stationary equations lies outside the region, should it be included as an interior candidate? Explain.
- For \(f(x,y)=x^2+y^2-2x+4y\), find the stationary point by solving both equations.
- Why does finding an interior stationary point not by itself prove that it is an absolute maximum or minimum on the region?